Two Theorems for Computation of Projections of Virtual Displacements and Its Application in Structural Analysis

Two Theorems for Computation of Projections of Virtual Displacements and Its Application in Structural

Analysis

Zhenchao Su

¹

, Yanxia Xue

^1*

Received 23 August 2016; accepted after revision 27 November 2016

Abstract

A theorem for planar case and its generalization for spatial case are proposed to determine the projection of a virtual dis- placement to the orientation under the case of knowing the projections of a virtual displacement to the given two or three orientations for object systems subject to holonomic and scle- ronomic constraints. Some lemmas corresponding to the two theorems for special cases are given. Applications to structural static analysis are investigated using the two theorems in this paper. Result reveals that the two theorems and corresponding lemmas are easy to be used, shorten the distance between the principle of virtual displacement and its application, and the relating problems can be solved quickly with them.

Keywords

projection of virtual displacement, principle of virtual dis- placement, engineering mechanics, structural analysis

1 Introduction

It is well known that the principle of virtual displacement (or virtual work) is a main part in analytical statics, and the import- ant basis for analytical dynamics and structural analysis. This principle provides an excellent tool for people to investigate the equilibrium laws of object systems, and plays a supporting role in classical mechanics.

Generally speaking, some current college textbooks [1-7]

for engineering mechanics or structural analysis course often propose two main methods for building relations among virtual displacements of different points for the system with holonomic and scleronomic constraints, ie, analytical method and geomet- rical method. Using analytical method, one can determine the virtual displacements of interest points (usually points of forces) by taking differentials of their position coordinates, and geomet- rical methods by introducing the concept of the instant center of rotation of virtual displacements. These methods are very useful in application of the principle of virtual displacement. However, in actual experience, we realize that these methods are some- times not sufficient for analyzing complex structures. For some difficult problems, such as the examples given in this paper, it is impossible or not sufficient to solve them out if only directly using the results given in these current textbooks. Then, how to shorten the distance between the principle of virtual displace- ment and its application, and give some feasible approaches to realize, is very important and great significance.

In this study, we will gain an insight into the application of the principle of virtual displacement and present two new the- orems on how to build up the relations among the projections to different orientations of the virtual displacement of a point.

Several examples are given to illustrate the application of two theorems and corresponding lemmas. The article is organized as follows. In Section 2, a theorem is given to find a projection under knowing the two projections to different orientations of a point for planar systems, and two lemmas for special cases are proposed. In Section 3, a theorem for spatial systems as the planar extending situation is given to find a projection under knowing the three projections to different orientations of a point, and three lemmas for special cases are proposed.

1 Department of Civil Engineering, Tan Kah Kee College,

Xiamen University, 363105, Zhangzhou city, Fujian Province, China

* Corresponding author, e-mail: xueyx888@126.com

61(2), pp. 135-139, 2017 https://doi.org/10.3311/PPme.9922 Creative Commons Attribution b research article

PP Periodica Polytechnica

Mechanical Engineering

2 The first theorem for plane cases

For a planar object system, if the projections of a virtual displacement dr of a certain dot to orientations n₁ and n₂ are dr1

anddr2 separately, shown in Fig. 1, then the projection of the virtual displacement to orientation n₃ is

δ δ θ φ δ φ

r r θ r

3

1 2

= ^sin

(

−

)

^{+ sin}

sin ,

where, angles θ and φ are angles between n₁ and n₂ , n₁ and n₃ separately.

Fig. 1 A virtual displacement and its projections

Proof. Assume e, e₁, e2 and e₃ are the unit vectors of the orientations of n, n₁, n₂ and n₃, separately. Take the start point A of the virtual displacement δr as the origin of the coordinate system xAy. Let a is the angle between n₁ and x axis. According to the given conditions, one reads

δr ∙ e₁ = δr₁ , δr ∙ e₂ = δr₂ . Hence, one can get

δr⋅

(

^cosαi+^sinαj

)

⁼δr1

δr⋅_^cos

(

^{α θ}+

)

i^+sin

(

^{α θ+}

)

j_{ =}δr2

where i, j are the standard unit vectors of x and y axis.

By taking the inner products δr ∙ i and δr ∙ j as unknowns, and solving the equations of (2) and (3), one can get

δ δ δ

r i⋅ = r1sin

(

+

)

⁻ r2sin sin

α θ α

θ

δ δ δ

r j⋅ = r2cos − r1cos

(

+

)

sin

α α θ

θ

Then, the projection of the virtual displacement to orienta- tion n₃ is

δ δ δ

δ r

r

3 3

1

= ⋅ = ⋅_

(

+

)

⁺

(

⁺

)

_

=

(

+

)

^⋅

(

⁺

)

⁻

r e r cos i sin j

cos sin

α φ α φ

α φ α φ δδ

δ δ

δ

r

r r

r

2

2 1

1

sin sin

sin cos cos

sin sin

α θ

α φ α α φ

θ α φ

+

(

+

)

^{⋅ −}

(

⁺

)

=

(

+

)

^{+ δδ}r2sin

sin φ.

θ

Lemma 1. When θ ≠ 0(180°), δr1 = δr₂ = 0, then δr₃ = 0. Lemma 2. When θ = 90°, then δr3 = δr₁ cosφ + δr₂ sinφ.

For demonstrating the effectiveness of this theorem in solving the balancing problems of planar object systems, two examples are given below.

Example 1. Determine the force in member CD of the truss, shown in Fig. 2. Assume all members are pin connected, and AD = AE = ED = EC = CG = DG = DB = BG = a.

Fig. 2 Example 1

Solution. To calculate the force in rod CD, it is isolated from the system in Fig. 2 (b). Then, the system in Fig. 2 (b) is a mechanism. Obviously, equilateral triangles AED and DBG can be regarded as rigid plates, and DAED can be assumed to rotate about A point with δθ. Then, δr_E = δr_D = aδθ. Con- sidering the orientation of δr_D and character of support B, point B is the virtual displacement center of DDBG, therefore δr_G = δr_D = aδθ. Based on the principle of virtual displace- ment, one reads

F r⋅δ_G+ ′ ⋅F_CD δr_D+F_CD⋅δr_C=0

Based on the theorem of projection of virtual displacement (i.e. the projections of the virtual displacements of the points from a straight line belonging to a body, on that line, are equal.), the projections of δr_C to the orientations E→C and G→C are zeros. Then, by employing the lemma 1 of the above theorem, δr_C must be zero. Therefore, based on the above analysis and (6), one can get

− ⋅F aδθ⋅cos30° − ′ ⋅F_CD aδθ =0. Because of θ ≠ 0, then

′ = = − ⋅ ° = − F_CD F_CD F cos30 3F.

2 (bar CD in compression).

Example 2. Determine the force in member DG of the truss, shown in Fig. 3. Assume all members are pin connected.

(1)

(3) (2)

(5) (4)

(6)

Solution. To calculate the force in rod DG, it is isolated from the system in Fig. 3 (b). Obviously, triangles ACE and DBE can be regarded as rigid plates, and DACE can be assumed to ro- tate about A point with δθ. Then, δr_C=4aδθ,δr_E =4 2aδθ. Considering that the orientation of δr_B is horizontal, the virtual displacement δr_D must be horizontal, and

δr_D =δr_E⋅cos45° =4aδθ, δr_B⋅cos45° =δr_E, δr_B=8aδθ. For the bar BH, it is easy to get that

δr_H⋅cos

(

90° −2α

)

=δr_B⋅cosα

δr δr a δ

H = 1 B=

2

4

sin sin .

α α θ

Then, for the bar GH, the projection of δr_G to the orienta- tion G→H is

δr δr a δ

GH = H⋅cos = cos

sin .

α α

α θ

4

For the bar CG, the projection of δr_G to the orientation C→G is

δrCG=δrC⋅cos

(

90° −β

)

=4asinβ θδ .

For the point G, the projections δr_GH and δr_CG are known.

By employing the formula (1) of the above theorem, the projec- tion of δr_G to the orientation G→D is

δ δ δ

r r r

GD

GH CG

=

{

−_ +

(

° −

)

_

}

⁺  ⁺

⁽

^{° −}

⁾



= −

sin sin

sin

β β β β β

β

2 90 2 90

4

4 2

4

8 4

a a

a a

cos sin

sin sin sin

cot cos sin

α β

α β β θ

α β β θ

 +

 



= −

(

+

)

δ δ

== − ⋅ ⋅ + ⋅

 

 = −

8 4 2

2 5 4 4

2 5

24

a a δθ 5aδθ.

By employing the principle of virtual displacement, one can get F r⋅δ_D+F_DG⋅δr_D+ ′ ⋅F_DG δr_G=0

Therefore, based on the above analysis and (7), one can get

F a⋅⁴ −FDG⋅⁴a ⋅

(

⁹⁰° −

)

^{− ′ ⋅}FDG ²⁴a ⁼

5 0

δθ δθ cos β δθ .

Because δθ ≠ 0, FDG = ′FDG, then

F_DG= F F F F

+ =

+ = =

sinβ ⁶ 5

2 5

6 5

5

8 0 279. .

3 The second theorem for spatial cases

If the projections of the virtual displacement δr of a dot to three non-coplanar orientations n₁ , n₂ , n₃ are δr₁ , δr₂ , δr₃ , then the projection δr₄ of the virtual displacement to the orientation n₄ is

δ

δ

r e e e

e e e

e e e

e e e

r

x y z

x y z

x y z

x y z

4 4 4 4

1 1 1

2 2 2

3 3 3

1

=

( )

^















−

, ,

1 1 2 3

δ δ r r















.

where e_ix ,e_iz ,e_iz are projections of unit vector e_i of spatial ori- entation n_i to the Cartesian coordinate axes x, y, z, separately.

(i = 1,2,3,4)

Proof. Assuming that i, j, k are the standard unit orthogonal vectors of the spatial coordinate system, based on the given conditions, one reads

e_i=e_ixi+e_iyj+e_izk , i=1 2 3 4, , , and

e e e r

e e e r

e

x y z

x y z

x

1 1 1 1

2 2 2 2

3

δ δ δ δ

δ δ δ δ

δ

r i r j r k

r i r j r k

⋅ + ⋅ + ⋅ =

⋅ + ⋅ + ⋅ =

rr i⋅ + r j⋅ + r k⋅ =





 e3yδ e3zδ δr3

or

e e e

e e e

e e e

x y z

x y z

x y z

1 1 1

2 2 2

3 3 3

















⋅

⋅

⋅













 δ δ δ

r i r j r k=















 δ δ δ r r r

1 2 3

.

Because the three orientations e₁,e2,e3 are non-coplanar, the coefficient matrix is reversible. Hence, one can get

δ δ δ

r i r j r k

⋅

⋅

⋅















=











 e e e 

e e e

e e e

x y z

x y z

x y z

1 1 1

2 2 2

3 3 3



















-1 1 2 3

δ δ δ r r r

. Fig. 3 Example 2

(7)

(8)

Thus, the projection of the virtual displacement δr to ori- entation n₄ is

δ

δ δ δ

r e e e

e e e

e e

x y z

x y z

x

4 4 4 4

4 4 4

1 1

=

( )

^⋅⋅

⋅

















=

( )

, ,

, ,

r i r j r k

yy z

x y z

x y z

e

e e e

e e e

r r r

1

2 2 2

3 3 3

1 1 2 3

































− δ

δ δ

.

Lemma 3. If the unit vectors e_i (i = 1,2,3,4) are normal orthogonal unit vectors, the projection δr₄ of the virtual dis- placement to the orientation n₄ is

δ

δ r e e e δ

e e e

e e e

e e e

r

x y z

x x x

y y y

z z z

4 4 4 4

1 2 3

1 2 3

1 2 3

1

=

( )

^















, , rr

r

2

δ3















.

Lemma 4. If the unit vectors e_i of orientations n_i (i = 1,2,3,4) are non-planar vectors, and the projections of the virtual dis- placement of a dot to n_i are all zeros, then the virtual displace- ment must be zero vector, and projection of the virtual displace- ment to n₄ (every orientation) must be zero.

Lemma 5. If the unit vectors e_i (i = 1,2,3,4) are standard unit orthogonal vectors of the Cartesian coordinate axes x, y, z, separately, the projection of the virtual displacement to e_i is δr_i , then the projection of the virtual displacement to the orientation n₄ is

δr e r e r e r₄ = _4xδ₁+ _4yδ₂+ _4zδ₃.

Example 3. Determine the force in member AC of the spatial truss, shown in Fig. 4 (a). Assume all members are pin con- nected. F₁=5kN, F₂=4kN, F₃=2kN, F₄=1.5kN, F₅=1.5kN.

Solution. To calculate the force in rod AC, it is isolated from the system in Fig. 4 (b). Obviously, hinge points E, D, C and B are fixed. Therefore, the projection of point A to the orienta- tions B→A and D→A are zeros. Assuming that the projection along the F₄ direction is dr, considering that the unit vectors of the direction B→A, A→C, D→A and F₄ are as follows

e_BA= − e_AC

 

 = −

( )

4 5

3

5 0 1 0 0

, , , , , ,

e_DA=

(

^{0 0 1, ,}

)

^, e_F4 ⁼

(

^{0,−1 0,}

)

Then, utilizing the second theorem for spatial case, the pro- jection along the A→C direction is

δ

δ r

r

4

1

1 0 0 4 5

3 5 0

0 0 1

0 1 0

0 0

1

= −

( )

−

−





































= −

−

, ,

,, ,0 0 .

5

4 0 3

4

0 0 1

0 1 0

0

0 3

( )

4

−

−



































= δ

δ r

r

By employing the principle of virtual displacement, one reads

FAC⋅δr F rA+ ₄⋅δA=0.

Based on the above analysis and (11), one can get F_AC⋅3 r F r+ ⋅ =

4δ ₄ δ 0

Because δθ ≠ 0, then one can get FAC= −4F = − × = −

3

4 3 1 5

4 . kN 2.0kN.

Fig. 4 Example 3

(9)

(10)

(11)

a)

b)

4 Conclusions

This paper mainly faces to the difficulty of computation of virtual work, investigates on the projection of a dot’s virtual displacement to a given orientation, propose two theorems and corresponding lemmas, and discuss their application for ana- lyzing the forces of structural members. Computation proces- sion reveals that the formula for planar situation is easy to use, and the formula for spatial situation is normalized and easy to be remembered. Combining with current methods, these results will be very helpful to analyze inner forces of complicit struc- tures. It’s necessary to say that the two theorems can be used in dynamics when dynamical problems convert into static prob- lems in form by utilizing D’ Alembert’s principle.

Acknowledgement

This paper is supported by the 13^th five-year educational scientific research planning programs in Fujian (Grant No.

FJJKCGZ16-152). The authors gratefully acknowledge its financial supports.

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